\documentclass[11pt,letterpaper]{article}

\newcommand{\mytitle}{Math120 Homework 1}
\newcommand{\myauthor}{Kevin Lewi}
\date{October 6, 2011}

\usepackage{hwformat}

\begin{document}

\maketitle

\section{Exercise 1.1.9}

\subsection*{Part A}

For closure, if $x$ and $y$ are in $G$, then let $a,b,c,d \in \mathbb{Q}$ be 
such that $x = a + b \sqrt{2}$ and $y = c + d \sqrt{2}$. Then, $x+y = (a+c) + 
(b+d) \sqrt{2}$. Since addition under $\mathbb{Q}$ is closed, then $x+y \in G$.

Associativity of $G$ follows from the associativity of addition under 
$\mathbb{Q}$.

The identity is $0$ for $G$, and it is unique.

For $x = a + b \sqrt{2}$, the inverse of $x$ is $-a - b \sqrt{2}$, and it is 
unique.

\subsection*{Part B}

easy

\section{Exercise 1.1.22}

neither part is hard.

\section{Exercise 1.1.32}

very easy.

\section{Exercise 1.2.5}

easy, use the generator presentation to show it.

\section{Exercise 1.2.9}



\end{document}
